Sulfur dioxide, SO₂, can be both an oxidising agent and a reducing agent because sulfur is in the intermediate oxidation state +4. With a sufficiently strong reducing agent such as hydrogen sulfide, sulfur in SO₂ can be reduced to elemental sulfur, accepting electrons. With a suitable oxidising agent such as bromine or acidified permanganate, sulfur in SO₂ can be oxidised to sulfate, donating electrons. The chemical role depends on which reactant it meets.
For JC students searching for an H2 Chemistry tutor in Hougang, SO2 oxidising and reducing agent, sulfur dioxide hydrogen sulfide equation, SO2 bromine water decolourisation, acidified KMnO4 redox half-equations, oxidation number sulfur +4 or A-Level Chemistry 9476, this is a useful way to practise electron accounting rather than memorise apparently contradictory labels.
The two examples to know immediately are:
- SO₂ acts as an oxidising agent with H₂S: SO₂ + 2H₂S → 3S + 2H₂O. Sulfur in SO₂ falls from +4 to 0.
- SO₂ acts as a reducing agent with Br₂: SO₂ + Br₂ + 2H₂O → SO₄²⁻ + 2Br⁻ + 4H⁺. Sulfur rises from +4 to +6.
- The rule: an oxidising agent accepts electrons and is reduced; a reducing agent donates electrons and is oxidised.
- What does not change: sulfur dioxide remains SO₂ at the start of both reactions, but its reaction partner and final sulfur species differ.
A student’s first difficulty is often not balancing the full equation; it is the meaning of the word agent. The oxidising agent causes something else to be oxidised while undergoing reduction itself. Once this is secure, the two roles of SO₂ become consistent.
The 2027 Singapore H2 Chemistry 9476 syllabus includes oxidation states, balancing redox equations, relevant inorganic reactions and applying electron-transfer reasoning. Sulfur dioxide in industrial and atmospheric contexts provides a useful cross-topic application; the exact full list of SO₂ reactions in this guide should not be assumed compulsory recall beyond the official examination-year syllabus.
At eduKate, small-group instruction of no more than three students emphasises an independently explained changed problem after repairing the first incorrect step. Families reading from Hougang should confirm actual lesson venue, current availability, teaching times and travel arrangements directly; the title does not establish a physical classroom in Hougang.
Hougang Chemistry learning library · 2027 H2 Chemistry syllabus directory
First, what is an oxidation state?
An oxidation state is a formal electron-accounting number assigned to an atom in a chemical species according to defined conventions.
It is not necessarily an experimentally measured electrical charge located entirely on that atom. In molecules with covalent bonds, electrons are shared and the formal oxidation number simplifies that complex distribution.
For many common sulfur compounds, the conventions oxygen = −2 and hydrogen = +1 are enough to determine sulfur’s oxidation state.
For neutral SO₂, let the sulfur oxidation state be x:
x + 2(−2) = 0.
So x = +4.
This is the key starting fact. Sulfur in SO₂ is not at the highest possible common oxidation state associated with sulfate, +6, and it is not at the low oxidation state −2 found in H₂S.
It occupies an intermediate position in the familiar sulfur oxidation-state range.
Therefore it has opportunities to increase or decrease its formal oxidation state under suitable reaction conditions.
The three useful sulfur oxidation states
Consider three substances:
H₂S: sulfur oxidation state −2.
S in elemental sulfur: oxidation state 0.
SO₂: sulfur oxidation state +4.
SO₄²⁻: sulfur oxidation state +6.
The sequence makes the electron-transfer logic visible.
Going from +4 to +6 is an increase in oxidation number, so sulfur is oxidised.
Going from +4 to 0 is a decrease in oxidation number, so sulfur is reduced.
That does not mean a change in oxidation number is the only criterion needed to predict which reaction happens. The reaction partner, thermodynamic driving force, kinetics and conditions still matter.
But it is the correct way to classify electron transfer in a reaction once the reactants and products are known.
Students who can locate +4 between 0 and +6 can quickly recognise why the sulfur dioxide examples have opposite redox roles.
What is an oxidising agent?
An oxidising agent is a substance that causes another species to lose electrons.
It does this by accepting electrons, which means that the oxidising agent undergoes reduction.
In a general equation:
Oxidising agent + electrons → reduced form.
For example, in the acidic reduction:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O,
permanganate receives electrons and serves as an oxidising agent toward a suitable reductant.
When SO₂ meets hydrogen sulfide, SO₂ can act as the oxidising agent because its sulfur centre accepts electron equivalents and reaches a lower oxidation state.
The common mistake is to write that an oxidising agent must itself be oxidised.
It is the opposite: the agent gets reduced while oxidising its partner.
What is a reducing agent?
A reducing agent is a substance that causes another species to gain electrons.
The reducing agent donates those electrons, so the reducing agent itself undergoes oxidation.
In a general equation:
Reducing agent → oxidised form + electrons.
For example, iodide can donate electrons and be oxidised to iodine:
2I⁻ → I₂ + 2e⁻.
Sulfur dioxide can serve as a reducing agent when a strong oxidant such as bromine accepts electrons from it.
The sulfur in SO₂ increases from +4 to +6 as it becomes a sulfate species.
Again, the words describe what a substance does to its partner.
A species that donates electrons is a reducing agent, even if its name contains oxygen and sounds like something that should always be an oxidant.
How SO₂ oxidises hydrogen sulfide
Hydrogen sulfide, H₂S, contains sulfur at oxidation state −2.
Sulfur dioxide contains sulfur at +4.
Under suitable reaction conditions, these species can react to form elemental sulfur and water:
SO₂ + 2H₂S → 3S + 2H₂O.
The two H₂S molecules contribute two sulfur atoms, and SO₂ contributes one. Thus the right side contains three sulfur atoms.
The hydrogen atoms are balanced as 4H on each side, and the oxygen atoms are balanced as 2O on each side.
This equation describes the familiar net chemistry of the Claus reaction, an important sulfur-recovery process. Industrial treatment requires controlled catalysts, temperatures and process stages; the equation is not a home-experiment procedure.
The US Environmental Protection Agency’s sulfur oxides reference identifies this reaction as an example of SO₂ behaving as an oxidising agent with H₂S.
The redox roles follow from the changes in oxidation state, not from the colour of the products.
Track both sulfur sources separately
Take:
SO₂ + 2H₂S → 3S + 2H₂O.
The sulfur from SO₂ changes:
+4 → 0.
That sulfur is reduced.
Each sulfur from H₂S changes:
−2 → 0.
Those sulfur atoms are oxidised.
The overall reaction therefore contains both oxidation and reduction.
SO₂ causes H₂S to be oxidised, so SO₂ is the oxidising agent.
H₂S causes the sulfur in SO₂ to be reduced, so H₂S is the reducing agent.
A student who simply says “sulfur becomes sulfur, so there is no oxidation” has ignored that sulfur atoms began in different compounds and oxidation states.
It is possible for the same element to be present in both the oxidised and reduced reactants.
The correct method is to track sulfur from each starting species to the relevant product sulfur.
Derive the H₂S oxidation half-equation
Hydrogen sulfide is oxidised to elemental sulfur:
H₂S → S + 2H⁺ + 2e⁻.
Check hydrogen: two on both sides, as H₂S and 2H⁺.
Check charge: the left has charge zero; the right has +2 from protons and −2 from electrons, so the net charge is zero.
Sulfur changes from −2 to 0 and loses two electrons.
Because our full net equation contains two H₂S molecules, multiply this half-equation by two:
2H₂S → 2S + 4H⁺ + 4e⁻.
The electrons produced by H₂S will be accepted by sulfur dioxide in the corresponding reduction half-equation.
This is how the reaction balances charge and oxidation-number changes.
The half-equation is written in an acidic-form accounting convention; it describes net electron transfer, not necessarily a single collision involving freely isolated protons.
Derive the SO₂ reduction half-equation
Sulfur dioxide can be reduced to elemental sulfur under a suitable acidic-form bookkeeping model:
SO₂ + 4H⁺ + 4e⁻ → S + 2H₂O.
Check oxygen: SO₂ has two O; 2H₂O has two O.
Check hydrogen: 4H⁺ balances 4H in 2H₂O.
Check charge: +4 from protons and −4 from electrons cancel.
Sulfur changes from +4 to 0, a gain of four electrons.
Add this to the doubled H₂S oxidation half-equation:
2H₂S → 2S + 4H⁺ + 4e⁻.
The 4H⁺ and 4e⁻ cancel, giving:
SO₂ + 2H₂S → 3S + 2H₂O.
We have derived the overall reaction from two consistent electron balances.
That is more reliable than guessing coefficients for the final equation.
The role of SO₂ changes with a stronger oxidant
Now change the reaction partner.
Instead of hydrogen sulfide, consider bromine, Br₂, which can act as an oxidising agent in a suitable aqueous reaction.
The sulfur dioxide can now be oxidised to sulfate-related products.
Sulfur changes:
+4 in SO₂ → +6 in SO₄²⁻.
That is a loss of two electrons.
The bromine changes:
0 in Br₂ → −1 in Br⁻.
Each bromine atom gains one electron, for a total of two electrons per Br₂ molecule.
Therefore SO₂ acts as the reducing agent, supplying the electrons accepted by Br₂.
The same SO₂ formula has changed chemical role because the other half-reaction is different.
This is precisely why oxidising and reducing behaviour should not be treated as permanent labels assigned to a compound without reference to a reaction.
Balance SO₂ oxidation to sulfate
In acidic-form electron accounting:
SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻.
Check oxygen: 2 in SO₂ plus 2 in water gives 4; sulfate contains 4.
Check hydrogen: two water molecules supply 4H, appearing as 4H⁺.
Check charge: left zero; right −2 +4 −2 = 0.
Sulfur loses two electrons as its oxidation state rises from +4 to +6.
The bromine reduction half-equation is:
Br₂ + 2e⁻ → 2Br⁻.
Adding the halves gives:
SO₂ + Br₂ + 2H₂O → SO₄²⁻ + 2Br⁻ + 4H⁺.
The overall equation is balanced.
It is sometimes written in molecular acid form as:
SO₂ + Br₂ + 2H₂O → H₂SO₄ + 2HBr,
which is a convenient conceptual representation in water, where these acids are largely dissociated under ordinary dilute conditions.
Both presentations express the same net electron-transfer accounting.
Why bromine water can lose its colour
Aqueous bromine is typically described as yellow-orange or brown-orange at suitable concentrations.
If sulfur dioxide reduces Br₂ to Br⁻ under appropriate conditions, the molecular bromine amount decreases.
Bromide ions are colourless in the ordinary aqueous comparison.
Therefore the bromine colour can fade or disappear as Br₂ is consumed.
This provides evidence of a reaction, but it is not uniquely diagnostic of sulfur dioxide: other reducing substances and certain unsaturated organic compounds can also consume bromine.
The correct chemical explanation here is redox reduction of bromine to bromide, not electrophilic addition to a carbon–carbon double bond.
If the unknown sample is a gas, other confirmatory evidence may be needed.
A student should not say “bromine water decolourises, therefore the gas must be an alkene”. Sulfur dioxide provides an important counterexample to overusing the test.
A comparison with acidified permanganate
Acidified permanganate contains the strongly purple ion MnO₄⁻.
When it oxidises sulfur dioxide under a suitable sufficiently acidic system, the permanganate can be reduced to Mn²⁺, which is very pale at ordinary dilute concentrations.
Sulfur dioxide is oxidised to sulfate.
The overall net ionic equation can be written:
2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺.
Check sulfur: five SO₂ molecules produce five sulfate ions.
Check manganese: two permanganate ions produce two Mn²⁺ ions.
Check oxygen: left 2(4) + 5(2) + 2 = 20, right 5(4) = 20.
Check charge: left −2, right +4 −10 +4 = −2.
This is an example of sulfur dioxide acting as a reducing agent.
It can cause the purple permanganate colour to disappear under suitable conditions as MnO₄⁻ is consumed.
The specific reduction product depends on pH; do not use the Mn²⁺ equation indiscriminately in neutral or alkaline systems.
Why sulfur’s intermediate oxidation state matters
Sulfur has several familiar oxidation states, including:
- −2 in H₂S.
- 0 in elemental sulfur.
- +4 in SO₂ and many sulfite-related compounds.
- +6 in sulfate and sulfuric acid.
A species with sulfur at +4 has room, in oxidation-number terms, to be oxidised upward toward +6.
It also has room to be reduced downward toward 0 or other lower values.
But an intermediate oxidation state by itself does not prove every possible chemical reaction will occur readily.
We also need an appropriate reaction partner with a sufficient thermodynamic driving force and an accessible kinetic pathway.
Thus “SO₂ can be both” does not mean “SO₂ reacts instantly with everything”.
The full explanatory chain is:
Intermediate oxidation state → possibility of electron donation or acceptance → appropriate partner and conditions → actual balanced reaction.
That framework is useful for other intermediate-state reagents, including hydrogen peroxide and some transition-metal species.
Compare the two redox roles
| Reaction partner | Sulfur in SO₂ changes from | Electron behaviour of SO₂-derived sulfur | SO₂ role |
|---|---|---|---|
| H₂S | +4 to 0 | Gains four electrons per SO₂ | Oxidising agent |
| Br₂ in water | +4 to +6 | Loses two electrons per SO₂ | Reducing agent |
| Acidified MnO₄⁻ | +4 to +6 | Loses two electrons per SO₂ | Reducing agent |
The table connects oxidation number changes to the correct agent labels.
Students should be able to reconstruct the table from the half-equations rather than memorise the last column.
Worked example: theoretical sulfur formed from H₂S
Suppose a hypothetical mass-balance question states that 0.0150 mol SO₂ reacts completely with excess hydrogen sulfide according to:
SO₂ + 2H₂S → 3S + 2H₂O.
The stoichiometric ratio is:
1 mol SO₂ → 3 mol S atoms in the elemental sulfur product.
Therefore:
n(S) = 0.0150 × 3 = 0.0450 mol.
Using approximate Ar(S) = 32.1:
m(S) = 0.0450 × 32.1 ≈ 1.44 g.
The hydrogen sulfide required is:
n(H₂S) = 2 × 0.0150 = 0.0300 mol.
These are theoretical values under a specified complete reaction, not predictions of an industrial sulfur-recovery plant’s actual efficiency.
If H₂S were instead the limiting reagent, we would calculate product amounts from its available mole amount and the same balanced coefficients.
Worked example: bromine required to oxidise SO₂
Suppose 0.0200 mol SO₂ is completely oxidised by excess bromine under the simple net equation:
SO₂ + Br₂ + 2H₂O → SO₄²⁻ + 2Br⁻ + 4H⁺.
The SO₂:Br₂ mole ratio is 1:1.
Therefore the theoretical bromine requirement is:
n(Br₂) = 0.0200 mol.
The amount of sulfate product is also:
n(SO₄²⁻) = 0.0200 mol.
Bromide production is:
n(Br⁻) = 2 × 0.0200 = 0.0400 mol.
This reaction involves two electrons per SO₂, matching the two electron acceptances required to reduce one Br₂ molecule to two bromide ions.
The stoichiometry can be checked in two ways: from the balanced equation or from the electron counts.
That cross-check is useful when an exam changes the oxidising agent.
Worked example: permanganate amount from SO₂
For the acidic permanganate reaction:
2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺,
suppose the available amount of sulfur dioxide is:
n(SO₂) = 0.00500 mol.
The permanganate-to-sulfur-dioxide mole ratio is 2:5.
Therefore:
n(MnO₄⁻) = 0.00500 × (2/5) = 0.00200 mol.
If a hypothetical standardized permanganate solution has concentration 0.100 mol dm⁻³, the theoretical solution volume is:
V = n/c = 0.00200/0.100 = 0.0200 dm³ = 20.0 cm³.
This is a numerical practice problem, not an instruction to prepare or analyse toxic SO₂ mixtures.
The result follows the two-electron sulfur oxidation and five-electron manganese reduction, which are reconciled by multiplying the two half-equations to transfer ten electrons.
Why a colour change alone does not prove the rate law
A solution may lose purple permanganate colour as a reaction proceeds.
If it loses colour more quickly under one set of conditions, that suggests the relevant chemical transformation is proceeding faster, assuming comparable optical and experimental conditions.
But the colour trace alone does not establish which elementary step controls the reaction or the powers in the rate law.
A rate law needs suitable concentration–time or initial-rate data and mechanistic justification.
The overall equation contains coefficients 2 and 5 for permanganate and sulfur dioxide, but it does not mean the rate is automatically second order in permanganate and fifth order in SO₂.
Chemical kinetics and net stoichiometry give different kinds of information.
That is a useful link to the H2 topic of rate-determining steps and reaction mechanisms.
Students should not infer kinetic exponents directly from overall balanced equations unless the question specifies an appropriate elementary event.
SO₂ in the atmosphere: a real-world connection
Sulfur dioxide is an important atmospheric pollutant emitted from some combustion, industrial and volcanic sources.
It can undergo chemical reactions in air and cloud droplets to form other sulfur-containing species, including sulfate aerosols and acidic products.
The conversion of sulfur(IV) to sulfur(VI) compounds is an oxidation process.
That makes sulfur dioxide a reducing agent relative to the oxidising species participating in those atmospheric reactions.
The atmospheric chemistry can involve oxygen, photochemistry, radical reactions, dissolved oxidants and catalysts. There is no single reaction equation that describes every situation in an urban or rural atmosphere.
The US Environmental Protection Agency’s sulfur oxides discussion notes that SO₂ may act as an oxidant or reductant in laboratory chemistry while undergoing atmospheric transformation into higher-oxidation-state sulfur compounds.
This is an example of how H2 redox reasoning connects with environmental chemistry without treating a simple test-tube equation as the full atmospheric model.
Why sulfur dioxide is not simply a strong acid
SO₂ dissolves in water and forms an aqueous sulfur(IV) acid–base system often represented using sulfurous acid and hydrogensulfite or sulfite species.
Its acid–base chemistry is separate from its ability to act as a reducing or oxidising agent.
For example, an aqueous equilibrium can be represented schematically as:
SO₂(aq) + H₂O ⇌ H⁺ + HSO₃⁻, with the actual hydration and solution speciation dependent on conditions.
This represents proton-transfer chemistry, not a change in sulfur’s oxidation state.
By contrast, conversion of SO₂-derived sulfur(+4) to sulfate sulfur(+6) is redox.
A student should identify whether a question asks about pH, species in water or oxidation-number changes before choosing an explanation.
A substance can participate in acid–base chemistry and redox chemistry in different contexts, just as a hydrated metal ion can be acidic without its metal oxidation state changing.
Do not confuse SO₂ with SO₃
Sulfur trioxide has formula SO₃, and sulfur in it has formal oxidation state +6.
Calculate:
x + 3(−2) = 0, so x = +6.
That differs from sulfur dioxide’s +4.
SO₃ has important reactions with water and participates in sulfuric-acid formation under suitable conditions, but it does not have the same simple upward oxidation-number headroom as SO₂.
SO₂ and SO₃ also have different molecular geometries in the familiar VSEPR comparison.
SO₂ is bent, while isolated molecular SO₃ is trigonal planar.
These differences illustrate that changing the number of oxygen atoms affects oxidation state, bonding and chemical behaviour.
A student should not replace SO₂ with SO₃ in a balanced equation just because both names begin with sulfur.
The precise formula is part of the chemical explanation.
Why hydrogen sulfide is the reducing agent in the Claus equation
Hydrogen sulfide contains sulfur at −2, a relatively low formal oxidation state for sulfur.
In:
SO₂ + 2H₂S → 3S + 2H₂O,
the sulfur from H₂S becomes elemental sulfur at oxidation state 0.
That is oxidation.
The electrons released by the H₂S-derived sulfur are accepted in the reduction of SO₂-derived sulfur from +4 to 0.
Therefore H₂S is the reducing agent.
This conclusion is not based on which gas is more acidic or which has more hydrogen atoms.
The number of H atoms helps balance the half-equation, but the redox role is identified from the oxidation-number change at sulfur.
This is an especially useful correction for students who confuse hydrogen content, acid strength and reducing power.
Why a substance can have more than one role
Many substances participate in different chemical reactions in different ways.
Water can be an acid or a base in different proton-transfer systems.
Hydrogen peroxide can be an oxidising agent or a reducing agent depending on its reaction partner.
Sulfur dioxide likewise can accept or donate electrons in appropriate redox systems.
The correct question is not “Is SO₂ permanently an oxidising agent or permanently a reducing agent?”
It is:
Which atom changes oxidation state, which direction does that change go, and what partner supplies or accepts the electrons?
That three-part method is more robust than learning one-word labels for compounds.
It also prevents an unnecessary contradiction when a student encounters two valid textbooks describing apparently opposite redox roles.
A common error table
| Student statement | What went wrong | Repair question |
|---|---|---|
| SO₂ always acts as a reducing agent | Ignored its oxidising reaction with H₂S | What is sulfur’s product oxidation state? |
| SO₂ always acts as an oxidising agent | Ignored its oxidation to sulfate | Can sulfur increase from +4 to +6? |
| The oxidising agent must lose electrons | Reversed the agent definition | Does the oxidant accept or donate electrons? |
| SO₂ + H₂S has no redox because all sulfur becomes S | Failed to track different sulfur sources | What oxidation state did each sulfur start in? |
| SO₂ to sulfate is a four-electron oxidation | Miscounted oxidation-number change | How far is +4 from +6? |
| Bromine-water decolourisation proves an alkene | Ignored bromine’s reaction with reductants | Could SO₂ reduce Br₂ to Br⁻? |
| Net equation coefficients equal rate-law powers | Confused stoichiometry with kinetics | What rate data are supplied? |
| Any purple permanganate reaction gives Mn²⁺ | Ignored the pH-dependent manganese products | Which acid conditions are specified? |
Eight independent practice questions
- Calculate the oxidation state of S in SO₂, H₂S and SO₄²⁻.
- Write the balanced net reaction between SO₂ and H₂S forming elemental sulfur and water.
- Identify the oxidising agent and reducing agent in that reaction.
- Balance the oxidation half-equation of SO₂ to SO₄²⁻ in acidic-form bookkeeping.
- Combine SO₂ oxidation with Br₂ reduction.
- Calculate theoretical sulfur moles from complete reaction of 0.0100 mol SO₂ with excess H₂S.
- Calculate Br₂ moles needed to oxidise 0.00500 mol SO₂ in the simple bromine equation.
- Explain why neither the purple colour change nor the overall reaction coefficients directly determine the rate law.
Answers
1. SO₂ +4, H₂S −2, SO₄²⁻ +6. 2. SO₂ + 2H₂S → 3S + 2H₂O. 3. SO₂ is the oxidising agent, H₂S the reducing agent. 4. SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻. 5. SO₂ + Br₂ + 2H₂O → SO₄²⁻ + 2Br⁻ + 4H⁺. 6. 0.0300 mol S. 7. 0.00500 mol Br₂. 8. Colour is an observation and balanced coefficients give net mole ratios; kinetics needs separate evidence or a justified mechanism.
If a student gets Question 2 right but Question 3 wrong, revisit the definition of an agent before doing more numerical redox practice.
A seven-session JC1–JC2 revision route
- Session 1: calculate sulfur oxidation states in H₂S, S, SO₂ and sulfate.
- Session 2: identify oxidising and reducing agents from changes in oxidation state.
- Session 3: balance the SO₂/H₂S equation through half-equations.
- Session 4: balance SO₂ oxidation to sulfate and its reaction with bromine.
- Session 5: solve hypothetical mole calculations for elemental sulfur and bromide.
- Session 6: compare SO₂ and other intermediate-oxidation-state species such as H₂O₂.
- Session 7: solve an unfamiliar redox pair and explain what the data support about feasibility, products and reaction rate.
The last session checks whether the learner can change the reaction partner without changing the fundamental definitions.
What parents near Hougang can ask an H2 Chemistry tutor
Bring a marked redox question and ask where the first error occurred: oxidation number, electron direction, agent label, half-equation balancing, or an unsupported inference from colour.
A useful small-group or private tutorial should not simply have a student copy two equations. It should ask for one independent changed example in which the same compound takes a different chemical role.
At eduKate, three-pax groups are intended to provide individual feedback. Confirm actual teaching location, current 9476 teaching availability, lesson schedule and commuting needs directly; a Hougang-focused educational guide does not itself establish a physical branch in Hougang.
Frequently asked questions
Can SO₂ be both an oxidising and a reducing agent?
Yes. Sulfur at +4 can be reduced to lower oxidation states or oxidised to +6 under suitable reactions and conditions.
Why is SO₂ an oxidising agent with H₂S?
SO₂-derived sulfur is reduced from +4 to 0 by accepting electrons from H₂S-derived sulfur, which is oxidised from −2 to 0.
Why is SO₂ a reducing agent with bromine?
SO₂ sulfur is oxidised from +4 to +6, donating electrons that reduce Br₂ to Br⁻.
Does SO₂ decolourise bromine water?
It can under suitable aqueous conditions because it reduces bromine to bromide. This is not uniquely diagnostic of SO₂.
What is the sulfate oxidation half-equation?
SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ in acidic-form electron accounting.
Does a colour change show a reaction’s rate law?
No. The rate law must be determined from appropriate kinetic measurements or a justified mechanism.
Continue learning and sources
Why hydrogen peroxide can be both oxidising and reducing · Why Fe³⁺ oxidises iodide to iodine · Why acidified permanganate is self-indicating · Why SO₂ changes acidified permanganate colour · US EPA: sulfur oxides, oxidising and reducing behaviour · Hougang Chemistry learning library.
Safety and model note: Both SO₂ and H₂S are hazardous gases. The equations are for conceptual and examination learning; this is not an instruction to generate, mix or handle these gases. Stoichiometry assumes the reactions proceed as stated.